Bethany, who weighs 420N, lies in a hammock suspended by two ropes tied to two trees. The left rope makes an angle of 45 degrees with the ground, the right one makes an angle of 30 degrees. Find the tension of the left rope. Find the tension of the right rope.

Eq1: T1* Cos135 = -T2*Cos30 .

T1 = 1.22T2.

Eq2: T1*sin135 + T2*sin30 = -420*sin270.
0.707T1 + 0.5T2 = 420, Replace T1 with 1.22T2:
0.863T2 + 0.5T2 = 420,
T2 = 308 N.
T1 = 1.22 * 308 = 376 N.

Both Eqs are satisfied.

;

412N and -227N? It said -227N wasn't the right answer.

All angles are measured CCW from +x-axis.

T1*Cos135 = -T2*Cos45.
T1 = T2.

T1*sin135 + T2*sin45 = -420*sin270.
0.707T1 + 0.707T2 = 420,
Replace T2 with T1 and solve for T1:

The right rope has a tension of 310 N for sure. But when I try to input it into bobpursley's two equations I can't find a value that fulfills both equations? If you know one of the ropes tensions for sure how do find the left rope's tension?

Well, aren't those some tricky angles! Let's see if we can untangle this problem together.

First, let's find the tension of the left rope. To do that, we need to find the horizontal component of the tension. With a 45-degree angle, the tension will be divided between the horizontal and vertical components in equal parts. So, we use some trigonometry and find that the horizontal component of the tension is:

Tension_left * cos(45) = Tension_left * (√2/2) = (Tension_left * √2) / 2

Now, let's find the tension of the right rope. This time, we need to find the vertical component of the tension. With a 30-degree angle, the tension will be divided between the vertical and horizontal components, but not in equal parts. To find the vertical component, we use more trigonometry:

Tension_right * sin(30) = Tension_right * (1/2) = (Tension_right) / 2

Since Bethany is at rest in the hammock, the vertical components of both tensions must balance out the weight of Bethany: 420N.

(Tension_left * √2) / 2 + (Tension_right) / 2 = 420

Now, I could solve this equation for you, but where's the fun in that? Why don't you give it a try? I'm sure you won't be "roped" in for too long!

To find the tension in the left rope, we can start by drawing a free body diagram for Bethany:

```
T1
/
|\
| \
| \
| \
| \
| \
| θ\
\|/____\
Bethany
```

In this diagram, T1 represents the tension in the left rope, and θ represents the angle made by the left rope with the ground. Since we are given the weight of Bethany (420N), we can draw a vector representing her weight (mg) acting vertically downwards. Then, using trigonometry, we can relate the weight to the tension in the left rope.

Let's break down Bethany's weight into its horizontal and vertical components:

Vertical component: mg * cos(θ)
Horizontal component: mg * sin(θ)

Since the hammock is in equilibrium, the sum of the vertical forces must be zero. We can set up the equation:

T1 * cos(θ) = mg * cos(θ)

Simplifying the equation, we can find the expression for T1:

T1 = mg

Given that Bethany's weight is 420N, we can substitute it into the equation to find the tension in the left rope:

T1 = 420N

Therefore, the tension in the left rope is 420N.

To find the tension in the right rope, we can follow the same process. Let's draw a free body diagram for Bethany again, focusing on the right rope this time:

```
T2
_______ <θ
\|/______\
Bethany
```

In this diagram, T2 represents the tension in the right rope, and θ represents the angle between the right rope and the ground.

Again, since the hammock is in equilibrium, the sum of the vertical forces must be zero. Setting up the equation:

T2 * cos(θ) = mg * cos(θ)

Simplifying the expression, we find:

T2 = mg

Substituting the weight (420N) into the equation, we can determine the tension in the right rope:

T2 = 420N

Therefore, the tension in the right rope is 420N.

horizontal equation: sume of forces is zero.

Tl*cos45-Tr*cos60=0
vertical equation: sume of forces is zero.
Tl*sin45+Tr*sin60-420=0

You have two equations, two unknowns. Solve